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  • I need help with A thin uniform, circular ring is rolling down an inclined plane of inclination 30° without slipping. Its linear acceleration along the inclined plane will be

A thin uniform, circular ring is rolling down an inclined plane of inclination 30° without slipping. Its linear acceleration along the inclined plane will be

  • Option 1)

    g/2

  • Option 2)

    g/3

  • Option 3)

    g/4

  • Option 4)

    2g/3

 

Answers (1)

best_answer

As we discussed in

Rolling of a body on an inclined plane -

a= \frac{g\sin \Theta }{1+\frac{K^{2}}{R^{2}}}

f= \frac{mg\sin \Theta }{1+\frac{R^{2}}{K^{2}}}

- wherein

K=Radius of gyration

\Theta = Angle of inclination

 

 

 a=\frac{MgSin\theta}{M+\frac{I}{R^{2}}} = \frac{Mg.Sin60\circ}{m+\frac{MR^{2}}{R^{2}}}

a=\frac{g}{2}


Option 1)

g/2

This is incorrect option

Option 2)

g/3

This is incorrect option

Option 3)

g/4

This is correct option

Option 4)

2g/3

This is incorrect option

Posted by

divya.saini

View full answer

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